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3 years ago

A ball is thrown at an angle of to the ground. If the ball lands 90 m away, what was the initial speed of the ball?

Physics

1 answer:

andre [41]3 years ago

8 0

Answer:

The initial speed of the ball is 30 m/s.

Explanation:

It can be assumed that the ball is thrown at an angle of 45 degrees to the ground. The ball lands 90 m away. We need to find the initial speed of the ball. We know that the horizontal distance covered by the projectile is called its range. It is given by :

$R=\dfrac{u^2\ sin2\theta}{g}$

u is the initial speed of the ball.

$v=\sqrt{\dfrac{Rg}{sin2\theta}}$

$v=\sqrt{\dfrac{90\times 9.8}{sin2(45)}}$

v = 29.69 m/s

v = 30 m/s

So, the initial speed of the ball is 30 m/s. Hence, this is the required solution.

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3 years ago

If the distance to a point source of sound is doubled, by what multiplicative factor does the intensity change?

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If the distance to a point source of sound is doubled, by a multiplicative factor of 4, the intensity changes.

Intensity of sound is the sound which is perpendicular to sound wave propogation per unit area. It is dependent on the Surface of source sound.

Intensity is the Power per unit area. Its SI unit is Watt/m².

As we move away from a source of sound, the sound starts to diminish. This is due to the decreasing sound intensity with distance.

It can also be understood by the fact that on increasing distance, the Power radiated by the source spreads over a larger area. Hence, the Intensity decreases gradually.

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1 year ago

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3 years ago

Read 2 more answers

Henry, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 830 N for the first 3.8 s after the eleva

Delicious77 [7]

Answer:

v= 4.0 m/s

Explanation:

When standing on the bathroom scale within the moving elevator, there are two forces acting on Henry's mass: Normal force and gravity.
Gravity is always downward, and normal force is perpendicular to the surface on which the mass is located (the bathroom scale), in upward direction.
Normal force, can adopt any value needed to match the acceleration of the mass, according to Newton's 2nd Law.
Gravity (which we call weight near the Earth's surface) can be calculated as follows:

$F_{g} = m*g = 95 kg * 9.8 m/s2 = 930 N (1)$

According to Newton's 2nd Law, it must be met the following condition:

$F_{net} = F_{g} -F_{n} = m*a\\ F_{net} = 930 N - 830 N = 100 N = 95 Kg * a$

As the gravity is larger than normal force, this means that the acceleration is downward, so, we choose this direction as the positive.

Solving for a, we get:

$a =\frac{F_{net} }{m} =\frac{100 N}{95 kg} = 1.05 m/s2$

We can find the speed after the first 3.8 s (assuming a is constant), applying the definition of acceleration as the rate of change of velocity:

$v_{f} = a* t = 1.05 m/s * 3.8 m/s = 4.0 m/s$

Now, if during the next 3.8 s, normal force is 930 N (same as the weight), this means that both forces are equal each other, so net force is 0.
According to Newton's 2nd Law, if net force is 0, the object is either or at rest, or moving at a constant speed.
As the elevator was moving, the only choice is that it is moving at a constant speed, the same that it had when the scale was read for the first time, i.e., 4 m/s downward.

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